Difficulty showing that a group G applied on X, G_{x}

Let ${\displaystyle G\curvearrowright X}$ be an action of a group on a set. For a point ${\displaystyle x\in X}$ the stabilizer of x is defined as ${\displaystyle G_x:=\{g\in G:gx=x\}}$. It is trivial that ${\displaystyle G_x}$ are subgroups of G: indeed, if g, ${\displaystyle h\in G_x}$ then since ${\displaystyle hx=x}$ we have ${\displaystyle x=h^{-1}x\text{so}g{h}^{-1}x=gx=x}$, and thus ${\displaystyle g{h}^{-1}\in G_x}$.
Now let ${\displaystyle x\in X}$ be a point and assume that for any ${\displaystyle y\in \text{or}b\left(x\right)}$ we have ${\displaystyle G_x=G_y}$. This of course means that ${\displaystyle G_x=G_{gx}\text{for all}g\in G}$. We show that ${\displaystyle G_x}$ is normal;
Let ${\displaystyle g\in G_x\text{and}h\in G}$. We have to show that ${\displaystyle hg{h}^{-1}\in G_x}$. But, note that ${\displaystyle G_{h^{-1}x}=G_x,\text{so}g\in G_{h^{-1}x}}$. We compute: ${\displaystyle hg{h}^{-1}x=h\left(g\left(h^{-1}x\right)\right)=h\left(h^{-1}x\right)=x,\text{so}hg{h}^{-1}\in G_x}$, proving normality.
Conversely, assume that ${\displaystyle G_x}$ is normal and let ${\displaystyle y\in \text{or}b\left(x\right),\text{say}y=gx}$ for some fixed ${\displaystyle g\in G}$. We need to show that ${\displaystyle G_x=G_{gx}}$. Let ${\displaystyle h\in G_x}$. Then ${\displaystyle g^{-1}hg\in G_x}$ by normality, so ${\displaystyle g^{-1}hgx=x,\text{so}hgx=gx,\text{so}h\in G_{gx}}$, showing that ${\displaystyle G_x\subset G_{gx}}$. Now let ${\displaystyle h\in G_{gx}}$; then ${\displaystyle hgx=gx,\text{so}{g}^{-1}hgx=x,\text{so}{g}^{-1}hg\in G_x}$. But since ${\displaystyle G_x}$ is normal, we also have ${\displaystyle g\left(g^{-1}hg\right)g^{-1}\in G_x,i.e.h\in G_x}$.

Step 1
Suppose ${\displaystyle G_x}$ is a normal subgroup. Then ${\displaystyle g^{-1}{G}_{x}g=G_x}$, from which we see that if ${\displaystyle g_y\in G_y\iff g^{-1}{g}_{y}gx=x}$. But ${\displaystyle g^{-1}{g}_{y}g}$ is thus in ${\displaystyle G_x}$ and thus also ${\displaystyle g{g}^{-1}{g}_{y}g{g}^{-1}=g_y\in G_x}$.
Also ${\displaystyle g_x\in G_x}$ if and only if for each g we get ${\displaystyle g_{x}gx=gx}$ (again because ${\displaystyle g^{-1}{g}_{x}gx=x}$), so ${\displaystyle g_x\in G_y}$.
Step 2
Now suppose ${\displaystyle G_x=G_y}$. Then we get ${\displaystyle g_{y}gx=gx\text{and}{g}_{y}x=x}$. Thus we get ${\displaystyle g_{y}gx=g{g}_{y}x}$ and thus ${\displaystyle g^{-1}{g}_{y}gx=x}$. Thus for each ${\displaystyle g_y\in G_y=G_x}$ we get ${\displaystyle g^{-1}{g}_{y}g\in G_x=G_y}$.
But I think this is as far as this goes now. We can get normality if we demand that ${\displaystyle G_x=G_y\text{for all}y\in G\left(x\right)}$.
EDIT: In fact we need this. Suppose ${\displaystyle g=e,\text{then}y=x}$. As ${\displaystyle G_x=G_x}$ it is easy to see that the given premisses are not suffiecient to get normality.